When 2 coins are tossed simultaneously What is the probability of getting at least one head?

When two coins are tossed simultaneously, all possible

outcomes are HH, HT, TH and TT.

 Total number of possible outcomes = 4

Let E be the event of getting at least one head.

    Then, the favourable outcomes are HT, TH and HH
    ​Number of favourable outcomes = 3

  ∴ P (getting at least 1 head)  = `"Number of favourable outcomes"/"Total number of possible outcomes "= 3/4`

Page 2

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

Let E be the event of getting an even number.

Then, the favourable outcomes are 2, 4 and 6.

Number of favourable outcomes = 3

∴ Probability of getting an even number =  P (E) = `3/6 = 1/2`

Page 3

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

Let E be the event of getting a number less than 5.

   Then, the favourable outcomes are 1, 2, 3, 4

   Number of favourable outcomes = 4

∴ Probability of getting a number less than 5 =  P (E) = `"Number of favourable outcome"/"Total number of possible outcomes"=4/6 = 2/3`

Page 4

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

Let E be the event of getting a number greater than 2.

Then, the favourable outcomes are 3, 4, 5 and 6.

Number of favourable outcomes = 4

∴ Probability of getting a number greater than 2 =  P (E) =`"Number of favourable outcomes"/"Total number of possible outcomes" = 4/6 = 2/3` 

Page 5

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

Let E be the event of getting a number between 3 and 6.

Then, the favourable outcomes are 4, 5

Number of favourable outcomes = 2

∴ Probability of getting a number between 3 and 6 =  P (E)= `"Number of favourable outcomes "/"Total number of possible outcomes" = 2/6 = 1/3`

Page 6

A die is thrown once. Find the probability of getting a number other than 3.

 Let E be the event of getting a number other than 3.  

   Then, the favourable outcomes are 1, 2, 4, 5 and 6.     

    Number of favourable outcomes = 5
∴ Probability of getting a number other than 3 =  P (E) = `"Number of favourable outcomes"/"Total number of possible outcomes" = 5/6`

Concept: Concept Or Properties of Probability

  Is there an error in this question or solution?

Page 7

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Total number of possible outcomes = 6

Let E be the event of getting the number 5.

Then, the favourable outcome is 5.

Number of favourable outcomes = 1

∴ Probability of getting the number 5 =  P (E) = `"Number of favourable outcome"/"Total number of possible outcomes" =1/6`

Text Solution

`1/4``1/2``3/4``3/8`

Answer : C

Solution : When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH, TT. <br> Total number of possible outcomes = 4. <br> Let E be the event of getting at least one head. <br> Then, E is the event of getting 1 head or 2 heads. <br> So, the favourable outcomes are HT, TH, HH. <br> Number of favourable outcomes = 3. <br> `:. ` P(getting at least one head ) ` =P(E) = 3/4`.